SOLUTION: Given n(U) = 26, n(A) = 11. n(B) = 12. and n(A ∩ B) = 8, find; a. n(A U B) b. n(B, but not A) I don't understand the concept of solving these. Please explain in a little

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Question 1189007: Given n(U) = 26, n(A) = 11. n(B) = 12. and n(A ∩ B) = 8, find;
a. n(A U B)
b. n(B, but not A)
I don't understand the concept of solving these. Please explain in a little depth on how to do this question.

Answer by math_helper(2461)   (Show Source): You can put this solution on YOUR website!

a)
If you draw a Venn diagram, you will have two overlapping circles, one for A and one for B. The region where they overlap is A ∩ B. There are 11 elements in A, 12 in B, and 8 in A ∩ B (so, some of the elements in A are in A ∩ B, and likewise for B). Thus n(A U B) = 12+11 -8 = 15 elements. The subtraction of 8 (= n(A ∩ B) ) is to account for the fact that adding n(A) and n(B) counts the intersection twice.
n(A U B) = n(A) + n(B) - n(A ∩ B)

Concrete Example:
----------------
Say A = { a, b, c, d, e, f, g }
and B = { a, e, i, o, u, y }
then A ∩ B = { a, e }
and A U B = { a, b, c, d, e, f, g, i, o, u, y }
n(A) = 7
n(B) = 6
n(A ∩ B) = 2
n(A U B) = n(A) + n(B) - n(A ∩ B) = 7 + 6 - 2 = 11 (which is easily verified by counting the number of elements in A U B directly)


b)
n(B, but not A) is the same as n(B) - n(A ∩ B) = 12 - 8 = 4


Referring to the example:
n(B) - n(A ∩ B) = 6 - 2 = 4 (4 elements of B are ONLY in B, and those elements are {i,o,u,y} )

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